Re: Filling the gaps in data arrays

[Mat086]

On 07/26/2018 10:47 AM, Mat086 wrote:
> I don't need to interpolate the data, actually the best thing would be to
> have NaN where the y-data doesn't match the X vector.
> this script is doing what I needed to do:
>
> dx = x(2)-x(1);
> Y = zeros(1,length([min(x):dx:max(x)]));
> X = [min(x):dx:max(x)];
> st=1;
> for k=1:length(X)
>    if isempty(y(x==X(k)))
>      Y(k) = 0;
>    else
>      Y(k) = y(st);
>      st=st+1;
>    end
> end
You still have the problem with accuracy; I think you will zero out some 
valid numbers because your calculated X values do not have to fall 
exactly on your x values.
For a contrived example, x=[0 0.3333333 0.666667 1]
x =

    0.00000   0.33333   0.66667   1.00000

octave:14> dx = x(2)-x(1);
octave:15> Y = zeros(1,length([min(x):dx:max(x)]));
octave:16> X = [min(x):dx:max(x)]
X =

    0.00000   0.33333   0.66667   1.00000

octave:17> x==X
ans =

   1  1  0  0

As you see, x and X are so very close that they print as identical, but 
they do not match. You can search for closeness:

my_eps=1e-6;
match = ( abs(x-X)<my_eps )

and then exploit the match vector to just transfer the matching y values:

dx = x(2)-x(1);
X = [min(x):dx:max(x)];
Y = zeros(1,length(X));
Y(find(match))= y

Octave/Matlab language has those 'vector' expressions that make it quite
fast and concise. The cost is maybe the extra storage (e.g. the match
vector, and the temporary vectors used by the find() operation) but in
practice it is a big win and it's good to learn to write code this way.

BTW, if you want to really use NaNs, you could do this:
Y = repmat(NaN,1,length(X));
&lt;/quote>


Dear Przemek,
many thanks for this, it was very helpful!

M



--
Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html